Q110 / 180 · COMEDK 2023 (Memory Based) · Mathematics
If $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$, where $x$ and $y$ are real, then the ordered pair $(2 x, 2 y)$ is
- A$(-6,0)$
- B$(0,6)$
- C$(0,-6)$
- Correct answerD$(1, \sqrt{3})$
Solution
$\begin{aligned} & \text { We have, }\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y) \\ & \Rightarrow \quad(\sqrt{3})^{50}\left(\frac{\sqrt{3}}{2}+i \frac{1}{2}\right)^{50}=3^{25}(x+i y) \\ & \Rightarrow \quad 3^{25}\left[-i\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)\right]^{50}=3^{25}(x+i y) \\ & \Rightarrow \quad(-i)^{50} \omega^{50}=x+i y \\ & \Rightarrow \quad\left(i^4\right)^{12} \cdot i^2 \cdot\left(\omega^3\right)^{16} \cdot \omega^2=x+i y \\ & \Rightarrow \quad(1)^{12} \cdot(-1) \cdot(1)^{16} \cdot\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=x+i y \\ & \Rightarrow \quad \frac{1}{2}+i \frac{\sqrt{3}}{2}=x+i y \\ & \Rightarrow \quad 1+i \sqrt{3}=2 x+i(2 y) \\ & \therefore \quad(2 x, 2 y)=(1, \sqrt{3})\end{aligned}$
Hence, the correct answer is option: 'D'.